- •Донецьк 2006
- •Integral calculus lecture no. 19. Primitive and indefinite integral
- •Point 1. Primitive
- •Properties of primitives
- •Point 2. Indefinite integral and its properties
- •Point 3. Integration by substitution (change of a variable)
- •Point 4. Integration by parts
- •Lecture no.20. Classes of integrable functions
- •Point 1. Rational functions (rational fractions)
- •Point 2. Trigonometric functions
- •Universal trigonometrical substitution
- •Other substitutions
- •Point 3. Irrational functions
- •Quadratic irrationalities. Trigonometric substitutions
- •Quadratic irrationalities (general case)
- •Indefinite integral: Basic Terminology
- •Lecture no. 21. Definite integral
- •Point 1. Problems leading to the concept ofa definite integral
- •Point 2. Definite integral
- •Point 3. Properties of a definite integral
- •I ntegration of inequalities
- •Point 4. Definite integral as a function of its upper variable limit
- •Point 5. Newton-leibniz formula
- •Point 6. Main methods of evaluation a definite integral Change of a variable (substitution method)
- •Integration by parts
- •Lecture no.22. Applications of definite integral
- •Point 1. Problem – solving schemes. Areas
- •Additional remarks about the areas of plane figures
- •Point 2. Arс length
- •Point 3. Volumes
- •Volume of a body with known areas of its parallel cross-sections
- •Volume of a body of rotation
- •Point 4. Economic applications
- •Lecture no. 23. Definite integral: additional questions
- •Point 1. Approximate integration
- •Rectangular Formulas
- •Trapezium Formula
- •Simpson10 formula (parabolic formula)
- •Point 2. Improper integrals
- •Improper integrals of the first kind
- •Improper integrals of the second kind
- •Convergence tests
- •Point 3. Euler г- function
- •Definite integral: Basic Terminology
- •Lecture no. 24. Double integral
- •Point 1. Double integral
- •Point 2. Evaluation of a double integral in cartesian coordinates
- •Point 3. Improper double integrals. Poisson formula
- •Point 4. Double integral in polar coordinates
- •Double integral: Basic Terminology
- •Differential equations lecture no.25. First and second order differential equations
- •Point 1. General notions
- •Point 2. Integrable types of the first order differential equations (of de - 1)
- •1. Separated de-1 (de-1 with separated variables)
- •2. Separable de-1 (de-1 with separable variables)
- •3. Homogeneous de-1
- •4. Linear de-1
- •5. Bernoulli de-1
- •Point 3. Order reducing second order differential equations
- •Lecture no.26. Second order linear differential equations
- •Point 1. General notions
- •Point 2. Linear dependence and independence
- •Point 3. Homogeneous equations Structure of the general solution of so lhde
- •So lhde with constant coefficients
- •Point 4. Nonhomogeneous equations Structure of the general solution of so lnde
- •Method of variation of arbitrary constants
- •Method of undetermined coefficients for so lnde with constant coefficients
- •Lecture no. 27. Systems of differential equations. Approximate integration of differential equations
- •Point 1. Normal systems of differential equations
- •Point 2. Approximate integration of differential equations Successive approximations method
- •Euler method
- •Differential equations: Basic Terminology
- •Bibliography textbooks
- •Problem books
- •Contents
Lecture no.22. Applications of definite integral
POINT 1. PROBLEM – SOLVING SCHEMES. AREAS.
POINT 2. ARС LENGTH
POINT 3. VOLUMES
POINT 4. ECONOMIC APPLICATIONS
Point 1. Problem – solving schemes. Areas
There are two schemes for finding some quantity Q.
1) Setting integral sum, expression Q as the limit of this integral sum, that is in the form of a definite integral.
Ex. 1. See three problems in P. 1 of the 21-th lecture and corresponding formu-las (10), (11), (12) in P. 2 of the same lecture.
2) Finding an element (in fact the differential) dQ of Q and then expression Q as a sum of all these elements. This procedure leads by the other way to a definite integral to evaluate Q.
To illustrate this second scheme we’ll consider the same three problems as in the lecture No. 21.
Ex. 2. The area of a curvilinear trapezium
(fig. 1).
Element (differential) dS
of the area S
(the area of a hatched strip with the base
on fig.1) equals
.
Adding all these elements from a
to b we
get the re-
Fig. 1 required area
as known definite integral
(
1 )
Ex. 3. Produced quantity U
of a factory during a time interval 0,
T.
Let
be a labour productivity of the factory. An element
of
the produced quantity du-
ring
an infinitely small time interval
equals
.
Adding all these elements from 0 to T we get the sought produced quantity U, namely
(
2 )
Ex. 4. The length path L traveled by a material point during a time interval from t = 0 to t = T with the velocity .
An element dL of the length path traveled during an infinitely small time interval is equal to
,
and the sum of all these elements from 0 to T gives the length path L to be found
.
( 3 )
Certainly, results (1), (2), (3) coincide with those (10), (11), (12) of the preceding lecture.
Additional remarks about the areas of plane figures
a) The case of nonpositive function. The area of a figure
,
represented of the fig. 2, equals
|
|
|
.
( 4 )
b) The case when a function has different
signs on various intervals. Let a
function
is nonnegative on the interval
and nonpositive on the interval
.
The area of a corresponding figure, represented on the fig. 3, equals
.
( 5 )
c) The case when a figure lies between two curves. The area of a figure
,
represented on the fig. 4, equals
.
( 6 )
Ex. 5. Find the area enclosed by two curves
(fig. 5).
Intersection
points of the curves have abscises
.
The figure is symmetric about the Oy-axis;
we may find double area of its right part.
Fig. 5
.
d) Figures oriented with respect to the Oy-axis. Areas of figures
(see fig. 6),
(see fig. 7),
(see fig. 8),
(see fig. 9),
are equal respectively
,
( 7 )
,
( 8 )
,
( 9 )
.
( 10 )
|
|
|
|
Ex.
6. Find the area of a figure bounded by curves
(fig. 10).
It’s well to write the equations of the curves in the next form
and to utilize the formula (10). Fig. 10 By virtue of symmetry of the figure with respect to the Ox-axis
.
e) The case of parametrically represented curve.
Let, for example, be given a curvilinear trapezium
, (fig. 1),
but
a curve
is determined by parametric equa-
Fig. 11
tions
(
for
and
for
).
To find the area of such the trapezium we change a variable
in the integral (1), namely
.
Ex. 7. Find the area of the loop of a curve
(fig. 11).
To construct the curve point by point and to see
the loop we equate to zero the expressions
,
and then form the next table
t |
-2 |
- |
-1 |
0 |
1 |
|
2 |
x |
4 |
3 |
1 |
0 |
1 |
3 |
4 |
y |
|
0 |
- |
0 |
|
0 |
- |
Point |
A |
B |
C |
O |
D |
E |
F |
From
the fig. 11 we see that
,
and so
.
f) The area in polar coordinates.
Let be given a curvilinear sector (or a curvilinear triangle) that is a plane figu-
|
|
|
re,
bounded by two rays
and a curve
given in polar coordinates
(fig. 12). Find the area of the curvilinear sector.An element
of the area is the area of a hatched circular sector with the radius
and a central angle
,
.
Adding
all these elements from
to
we get the area in question
.
( 11 )
Ex. 8. Find the area of a figure bounded by a
cardioid
7
(fig. 13).
Desired area equals the twofold area of the upper
part of the figure, which is a circular sector with the rays
.
.
Ex. 9. Find the area of a figure bounded by Bernoulli lemniscate8
.
The figure is symmetric with respect to the Ox-,
Oy-axes
and lies in the angles determined by the straight lines
(fig. 14). Passing to polar coordinates
,
we write the equation of the lemniscate in more suitable form
Then
we calculate the quadruplicated area of the circular sector bounded
by the lemniscate and the rays
.
We’ll obtain
.

Fig.
2
Fig.
3
Fig.
4
Fig. 6
Fig. 7
Fig. 8
Fig. 9
Fig. 12
Fig. 13
Fig. 14